inverter wattage to charge an x2?

[rotorblades]

one x2 what wattage do i need?i know it can vary and peaksometimes
im thinking 800 watts continuous for lithium-ion batteries,?

[jgbackes]

I use a 450 amp Cobra brand inverter

[rotorblades]

Originally Posted by jgbackes

I use a 450 amp Cobra brand inverter

seems pretty large but thanks or did you mean watts?

[jgbackes]

Yea Opps! I meant watts.

[Tritium]

If you plan on getting one for use in your car, you'd have to hook it direct to the battery.
However if you still plan on getting a SegVator lift some day, they come preinstalled on the lift.

The reason is that the 12volt cigarette plugs don't supply enough for the massive draw of the inverter/Segway

[rotorblades]

Originally Posted by Tritium

If you plan on getting one for use in your car, you'd have to hook it direct to the battery.
However if you still plan on getting a SegVator lift some day, they come preinstalled on the lift.

The reason is that the 12volt cigarette plugs don't supply enough for the massive draw of the inverter/Segway

yes thanks doyouknow if the standard trailerlight plug is usually wired okfor the segvator lift?

[rotorblades]

for an electronic device i wonder why seg is so secretive on how many charge watts. i know it varies ive read everything i can find i used to be an electronics student soo iwould thinkthat infot would be a little more available.especially in the manuals i glide on solar power= nuclear power direct to me.if i wasnt so laid up i would hook up my killa-watt meter up to see.

[Bob.Kerns]

Originally Posted by rotorblades

for an electronic device i wonder why seg is so secretive on how many charge watts. i know it varies ive read everything i can find i used to be an electronics student soo iwould thinkthat infot would be a little more available.especially in the manuals i glide on solar power= nuclear power direct to me.if i wasnt so laid up i would hook up my killa-watt meter up to see.

Well, if you do a forum search, you will find many, many discussions of this.

It's a little tricky to give a hard number, because the best answer depends on both the inverter and what your question is (inverter output vs input).

My observations with a Kill-a-Watt is that it doesn't go much over 120 W. But that doesn't give you much headroom for peak load, duty-cycle limitations of the inverter, inadequate wiring on the input leading to voltage variations, etc.

People report using 400W inverters successfully. You could probably get by with less -- 200W would likely work, if the unit is conservatively rated and properly wired. In between? We're in guesswork land.

[KSagal]

Originally Posted by Bob.Kerns

Well, if you do a forum search, you will find many, many discussions of this.

It's a little tricky to give a hard number, because the best answer depends on both the inverter and what your question is (inverter output vs input).

My observations with a Kill-a-Watt is that it doesn't go much over 120 W. But that doesn't give you much headroom for peak load, duty-cycle limitations of the inverter, inadequate wiring on the input leading to voltage variations, etc.

People report using 400W inverters successfully. You could probably get by with less -- 200W would likely work, if the unit is conservatively rated and properly wired. In between? We're in guesswork land.

Bob,

I have used inverters in the past, and what one company calls a modified sine wave may not look like what another calls one.

Typically when scoped, the wave from my cheap inverters have only a few large steps, making the AC wave look like a staircase up and back down. On the better inverters, the steps were much smaller, and there were more of them, making the AC wave look more like a pile of gravel.

I believe that RMS average wave values might list both of the above waves as the same voltage and frequency, but I wonder if the larger steps as opposed to the smaller ones provide less available current.

What think you?

[Bob.Kerns]

Originally Posted by KSagal

Bob,

I have used inverters in the past, and what one company calls a modified sine wave may not look like what another calls one.

Typically when scoped, the wave from my cheap inverters have only a few large steps, making the AC wave look like a staircase up and back down. On the better inverters, the steps were much smaller, and there were more of them, making the AC wave look more like a pile of gravel.

I believe that RMS average wave values might list both of the above waves as the same voltage and frequency, but I wonder if the larger steps as opposed to the smaller ones provide less available current.

What think you?

Well, the point of RMS (Root Mean Square) is to make the waveform irrelevant. Since higher voltages mean higher currents (and vice versa), the higher voltages end up being weighted more in the calculation. That's why the "square" in RMS. It's the square root of the average of the square of the voltage (or current).

That doesn't really answer your question, though. You're really asking about what it does *under load*. If the RMS voltages remained the same, they'd still be producing the same power output -- but will they remain the same as the load approaches the rated limit?

For that, we'd have to consider how the waveform is being produced. I haven't really looked at cheap inverters since the days when they used vibrating reed switches, so I don't know what they do to make them cheap, but having designed a few power supplies (LONG ago), I can tell you some of the tradeoffs.

The easiest, most efficient way to approximate a sine wave is to periodically turn on a switch to charge a capacitor up to the voltage it should be at at that point in time for a sine wave. This is efficient, because an ideal switch dissipates no power when it is on or off -- it only dissipates power when it's in the in-between state when resistances is neither zero nor infinite.

The switches (FETs, SCRs, Triacs, etc.) are not ideal switches, however. They have some resistance even when fully on, and they take some non-zero time to turn on and off.

A "chunkier" waveform will require more of the total charge to be passed in each chunk, and a bigger capacitor to store it. That means higher peak currents in charging, unless steps are taken to limit it (for example, putting a resister in series).

So bigger capacitor, higher peak currents, a (bigger) resistor in series wasting power -- that's not a design direction you want to go. I'd be surprised to find an inverter like that today!

Historically, I think they were designed like that because of the slow switching times of available high-powered solid-state switches. That's when most of the power is lost. Power lost = heat. So I think they had to limit the number of times they switched the power.

But how does this relate to your question? Well, the relationship is not simple. It depends on the design choices that were made. How big is the output capacitor, that stores the current while it's in the "off" state. Does the design turn it back on sooner when the voltage drops? (Probably not, since we're talking cheap&simple). How much is the current limited by the limits of the device? How big a heat sink is there, to dissipate heat? The thermal engineering is a critical part of designing power circuitry.

The device with the smoother waveform will have the same issues, though. It will switch more frequently, but will likely have a smaller output capacitor that drains more quickly under load. It will be using a modern device that switches more quickly, but it'll still be producing heat with each switch.

If the capacitor drained by the same percentage on each cycle for the two inverters, the RMS voltage would be affected the same. If each can bring the voltage back up to the same % of intended, likewise the RMS voltage would be affected the same.

But that's math & physics, not engineering. The engineering question is: can they?

So my *inexpert* expectation would be, that it would depend not so much on the waveform, but on how close the design pushes the limits for the devices in question.

And that's where price comes into the equation. I'd expect the cheaper device to push the limits more closely, be rated less conservatively, and perform more poorly.

But that's not necessarily true in individual cases. A clever designer and efficient manufacturer can produce a good product cheaply, while a sloppy design and inefficient manufacture can produce a poor product expensively -- and then spend a bunch on marketing, have a high markup, and drain your wallet.

I know, it's a non-answer. But I hope I at least answered your core question of how waveform interacts.